Bank Math

Rational number is any number that can express in the form of qp of two integers, where 'q' cannot be zero.

i) First rational number between 3 and 4 can be calculated by finding average between them, which is
(3 + 4)/2 = 7/2 or 14/4
Now, we have three numbers i.e. 3, 7/2 and 4, so other remaining rational numbers can be calculated by taking average between 3 and 7/2, and between 7/2 and 4.

ii) Second rational number between 3 and 7/2 can be calculated by finding average between them.
(7/2 + 3) / 2 = {(7 + 6)/2} / 2 = 13/4

iii) The third rational number between 4 and 7/2 can be calculated by finding the average between them.
(7/2 + 4)/2 = {(7 + 8)/2}/2 = 15/4

iv) Similarly fourth rational number between 3 and 13/4 can be calculated by finding the average between them.
(13/4 + 3) / 2 = {(13 + 12)/4}/2 = 25/8

v) Similarly, fifth rational number between 4 and 13/4 can be calculated by finding average between them.
{(13/4) + 4}/2 = {(13 + 16)/4} / 2 = 29/8

Then the rational numbers between 3 and 4 are : 7/2 or 14/4, 13/4, 15/4, 25/8, 29/8

Question: In a trapezoid, the lengths of the two parallel bases are 10 and 16. If the height of the trapezoid is 4, find the area of the trapezoid.

Solution: 
Given that, 
Trapezoid with bases a = 10 and b = 16
Height, h = 4 

We know, 
Area of trapezoid = (1/2) × (sum of bases) × height = (1/2) × (a + b) × h
= (1/2) × (10 + 16) × 4
= (1/2) × 104
= 52

So the area of the trapezoid is 52 square units.

Question: If an exponent or index has base 100 and power zero, then which of the following will be its value?
(Officer Cash 2022 অনুযায়ী)

Solution:
x⁰ = 1 (for any non-zero base a)

Now,
= 100⁰
= 1

Question: If x + y = a, x² + y² = b² and x³ + y³ = c³ then find the value of a³ + 2c³ = ? 

Solution: 
Given that, 
x + y = a .........(1)
x² + y² = b² .........(2)
And x³ + y³ = c³

Now, 
a³ + 2c³
= (x + y)³ + 2(x³ + y³)
= x³ + 3x²y + 3xy² + y³ + 2x³ + 2y³   ; [(a + b)³ = a³ + 3a²b + 3ab² + b³]
= 3x³ + 3y³ + 3xy(x + y)
= 3(x³ + y³) + 3xy(x + y)
= 3{(x + y)(x² - xy + y²)} + 3xy(x + y)  ; [a³ + b³ = (a + b)(a² - ab + b²)]
= 3(x + y)(x² - xy + y² + xy)
= 3(x + y)(x² + y²)
= 3ab² ; [From 1 and 2]

Question: A man bought 40 shares of Tk. 50 at 5tk discount, the rate of dividend being 27%. The rate of interest obtained is-

Solution:
Given,
Total investment = Tk. [40 × (50 - 5)] = Tk. 1800.
Face value = Tk. (50 × 40) = Tk. 2000.

∴ Dividend = (27/100) × 2000
= Tk. 540

∴ Interest obtained = (540/1800) × 100%
= 30%

In this type of question, We need to find out the LCM of the given numbers.
LCM of 12, 15, 18 and 20;
12=2 × 2 × 3;
15=3 × 5;
18=2 × 3 × 3;
20=2 × 2 × 5

Hence, LCM = 2 × 2 × 3 × 5 × 3

Since, the soldiers are in the form of a solid square.
Hence, LCM must be a perfect square. To make the LCM a perfect square, We have to multiply it by 5,
hence,
The required number of soldiers = 2 × 2 × 3 × 3 × 5 × 5
= 900

Average of 15 numbers = 15, Average of 5 numbers = 14, Average of 9 numbers = 16

Average = Total numbers/Number of Numbers
15 = Total numbers/15
Therefore, total numbers = 15 x 15 = 225.

Middle number = (Total numbers) – [(Average of 5 num x no of num) + ( Average of 9 num x no of num)]
= (225) – [(14 x 5) + (16 x 9)]
= (225) – (214)
= 11.

Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.
Then, the sum of their values = Tk. (25x/100) + (10 × 2x/100) + (5 × 3x/100) = Tk. 60x
Or, 60x/100 = 30
Or, x = (30 x 100)/60 = 50.
Hence, the number of 5 p coins = (3 x 50) = 150.

Let the missing number be x.
Given,
20 + 8 × 0.5/(20 - x) = 12
⇒ (20 + 4)/(20 - x) = 12
⇒ 24/(20 - x) = 12
⇒ 20 - x = 24/12
⇒ 20 - x = 2
⇒ x = 20 - 2
⇒ x = 18.

Question: A borrows sum of TK. 90000 for 4 years at 5% simple interest. He lends it to B at 7% for 4 years at simple interest. What is his gain?

Solution:
A's, Principal = TK. 90000
simple interest rate = 5%
time = 4 years

∴ simple interest = TK. (90000 × 4 × 5)/100
= TK. 18000

B's, Principal = TK. 90000
simple interest rate = 7%
time = 4 years

∴ simple interest = TK. (90000 × 4 × 7)/100
= TK. 25200

∴ Profit of A = TK. (25200 - 18000) = TK. 7200

Question: From P and Q, two trains start moving towards each other at the same time. Their speeds are 120 km/h and 100 km/h, respectively. When the two trains meet each other, one train has covered 40 km more than other train. Find the distance between P and Q?

Solution: 
Speeds are in the ratio 120 : 100 = 6 : 5
So distances covered in the same time are also in the ratio 6 : 5

Let distances be 6k and 5k.

∴ Difference = 6k - 5k = 40 
∴ k = 40

∴ Total distance = 6k + 5k = 11k = 11 × 40 = 440 km

So the distance between P and Q is 440 km.

Question: If A = {a, b, c, d, e, f, g} and B = {d, e, f, g}, then A - B = ?

Solution:
A - B = {a, b, c, d, e, f, g} - {d, e, f, g}
= {a, b, c}

As, R takes 40 days to complete the work, (P + Q) will take = 40/2 = 20 Days
R can do 1/40 part of the work in one day and (P + Q) can do 1/20 part of the work in a day.
So, in 1 day (P + Q + R) can do = (1/20 + 1/40) = (3/40) part of the work
∴ Days required to complete 1 or Total work is = 40/3 = 13(1/3) days

Question: A proficient worker is twice as efficient as an apprentice. After 10 days of joint work, they earn Tk. 45,000 together. What is the apprentice’s wage per day?

Solution: 
Given that, 
A seasoned workers efficiency is twice as much as an apprentice's.
They work together for 10 days
And earn Tk. 45000 together

Here, Efficiency of Seasoned Worker : Apprentice = 2 : 1
 Now, earning in 1 day = 45000/10 = Tk. 4500
So, Daily wage of Apprentice = 4500 × (1/3) = Tk. 1500

Thus, the daily wage of the apprentice is Tk. 1500.

Question: If A and B are in the ratio 3 : 4, and B and C are in the ratio 12 : 13. Then A and C will be in the ratio:

Solution:
দেওয়া আছে,
A : B = 3 : 4
⇒ A/B = 3/4

এবং
B : C = 12 : 13
⇒ B/C = 12/13

∴ (A/B) × (B/C) =  (3/4) × (12/13)
⇒ A/C = 9/13
∴ A : C = 9 : 13

Then A and C will be in the ratio is 9 : 13.

In Parallelogram
Let,
b and care sides
b = base
h = height
area = base × height
= b × h
Therefore,
p² - 4 = (p + 4)(p - 3)
p² - 4 = p² + p - 12
p = 8
Hence, actual area
= (p² - 4) = 8² - 4
= 64 - 4
= 60 square units.

Question: If log₂[log₃(log₂x)] = 1, then x is equal to = ?

Solution:
Given that, 
log₂[ log₃( log₂x)]=1
⇒ log₃(log₂x) = 2¹ = 2
⇒ log₂x = 3² = 9
⇒ x = 2⁹ = 512
∴ x = 512

Question: In how many years will Tk. 750 amount to Tk. 900 at 4% simple interest per annum?

Solution:
Simple Interest = Amount - Principal
= 900 - 750
= 150

Here,
Principal, P = 750
Interest Rate, R = 4%
SI = 150
Time, T = ?

SI = PRT/100
⇒  T = (SI × 100)/(P × R)
= (150 × 100)/(750 × 4)
= 15000/3000
= 5 years

প্রশ্ন: ফারহানের জন্মদিন 29 ফেব্রুয়ারি। তাঁর জন্ম গ্রহণের সালটি হতে পারে

সমাধান:
- ফেব্রুয়ারি মাস 29 দিনে হয় যখন বছরটি লিপ ইয়ার হয়।
- অপশনে প্রদত্ত সালগুলোর মধ্যে শুধুমাত্র 2008 হচ্ছে লিপ-ইয়ার।
- সুতরাং ফারহানের জন্মগ্রহণের সালটি হতে পারে 2008।

[গ্রেগরীয় অ্যালগরিদম অনুযায়ী, কোনো বছর ৪ দ্বারা বিভাজ্য এবং ১০০ দ্বারা বিভাজ্য না হলে তবে সেই বছর অধিবর্ষ হবে। কিন্তু ১০০ দ্বারা বিভাজ্য হলে সেই বছর অধিবর্ষ হবে না যদি ঐ বছর ৪০০ দ্বারা বিভাজ্য না হয়।]

Question: What should be the value of "Q" so that the expression (25 - 30x + Qx²) becomes a perfect square?

Solution:
(25 - 30x + Qx²)
= (5)² - 2 × 5 × 3x + (3x)² + Qx² - (3x)²
= (5 - 3x)² + Qx² - 9x²

∴ The expression becomes a perfect square if,
Qx² - 9x² = 0
⇒ Qx² = 9x²
∴ Q = 9

Question: Rafi weighs 72 kg. If he reduces his weight in the ratio 6 : 5, find his new weight​ in kg.

Solution:
ধরি, রাফির পূর্বের ওজন = 6x
রাফির পরের ওজন = 5x

প্রশ্নমতে,
6x = 72
⇒ x = 72 / 6 = 12

∴ ওজন কমে যাওয়ার পর হবে = 5x = 5 × 12 = 60 kg

Question: If y = √8 + √7, then what is the value of y³ + (1/y³)?

Solution:
দেওয়া আছে,
y = √8 + √7
⇒ 1/y = 1/(√8 + √7)
⇒ 1/y = (√8 - √7)/(√8 + √7)(√8 - √7)
⇒ 1/y = (√8 - √7)/{(√8)² - (√7)²}
⇒ 1/y = (√8 - √7)/(8 - 7)
∴ 1/y = √8 - √7

এখন, y + 1/y = (√8 + √7) + (√8 - √7)
= 2√8 = 2 × 2√2 = 4√2

এখন,
y³ + (1/y³)
= (y + 1/y)³ - 3(y)(1/y)(y + 1/y)
= (y + 1/y)³ - 3(y + 1/y)
= (4√2)³ - 3(4√2)
= (4³)(√2)³ - 12√2
= 64(2√2) - 12√2
= 128√2 - 12√2
= 116√2

সুতরাং, নির্ণেয় মান হলো 116√2

Question: If the sum of three consecutive numbers is multiplied by 4, the result is 132. What is the largest number?

Solution:
ধরি,
তিনটি ধারাবাহিক সংখ্যা যথাক্রমে x, x + 1 এবং x + 2

প্রশ্নমতে,
⇒ {x + (x + 1) + (x + 2)} × 4 = 132
⇒ (3x + 3) × 4 = 132
⇒ 3x + 3 = 132 / 4
⇒ 3x + 3 = 33
⇒ 3x = 33 - 3
⇒ x = 30/3
∴ x = 10

বৃহত্তম সংখ্যাটি হলো = x + 2
= 10 + 2
= 12

Increase = 7.5 - 6
= 1.5
percentage of increase= (1.5/6) × 100
= 25%
so, consumption must be reduced by (r/100+r) × 10
= (25/100+25) × 100
= 20%

Question: The simple interest at p% for p years will be Tk. p on a sum of?

Given,
Rate of interest, r = p%
Interest, I = p Tk.
Time, n = p years.

We know,
I = Pnr
P = I/nr
= (p × 100)/(p × p)
= 100/p Tk.

Question: A train running at a speed of 72 km/hr crosses a platform double its length in 45 seconds. What is the length of the platform in meters?

Solution: 
Let the length of the train be x meters
Then, length of the platform = (2x) meters

∴ Speed of the train = {72 × (5/18)} m/sec
= 20 m/sec

∴(x+2x)/20 = 45
⇒ 3x = 45 × 20
⇒ 3x = 900
⇒ x = 900/3 = 300

Hence, length of platform = 2x = (2×300)m = 600m

Question: A jar is filled with liquid, 2 parts of which are water and 4 parts juice. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half juice?

Solution:
Let the quantity of liquid in the vessel = 6 units.
Out of this liquid, x units are replaced by water.

Amount of water in the new mixture = {2 - (2x/6) + x} unit
= 2 - (x/3) + x
= (6 - x + 3x)/3
= (2x + 6)/3 unit

Amount of juice in the new mixture = 4 - (4x/6) unit
= 4 - (2x/3) 
= (12 - 2x)/3 unit

ATQ,
(2x + 6)/3 = (12 - 2x)/3
⇒ 2x + 6 = 12 - 2x
⇒ 4x = 6
⇒ x = 3/2 
 
∴ The fraction of the mixture that is replaced is = (3/2) × (1/6) 
= 1/4

Question: If log₃(x⁴ - x³) - log₃(x - 1) = 3 then x is equal to?

Solution: 
Given that, 
log₃(x⁴ - x³) - log₃(x - 1) = 3 
⇒ log₃[(x⁴ - x³)/(x - 1)] = 3
⇒ log₃[x³(x - 1)/(x - 1)] = 3
⇒ log₃x³ = 3
⇒ 3log₃x = 3
⇒ log₃x = 3/3 = 1
⇒ log₃x = 1
⇒ x = 3¹
∴ x = 3

Let the speed of Rizvi be x kmph;
Hence, Amin's speed = (x + 4) kmph;
Distance covered by Amin = 60 + 12 = 72 km;
Distance covered by Rizvi = 60 - 12 = 48 km.

According to question,
⇒ 72/(x + 4) = 48/x
⇒ 3/(x + 4) = 2/x
⇒ 3x = 2x + 8
⇒ x = 8 kmph.

Question: The average weight of 25 students in a class was found to be 48 kg. Later it was discovered that one student's weight was recorded as 35 kg instead of 60 kg. What is the correct average weight of the 25 students?

Solution:
Given,
The average weight of 25 students in a class was found to be 48 kg
Incorrect sum of the weight of 25 students = (48 × 25) kg
= 1200 kg

Correct sum of the weight of 25 students = (incorrect sum) - (wrongly copied item) + (actual item)
= (1200 - 35 + 60) kg
= 1225 kg

Correct mean = correct sum/number of students
= (1225/25) kg
= 49 kg

∴ Hence, the correct mean weight is 49 kg

Question: How many times are the hands of a clock at right angle in a day?

Solution:
In 12 hours, they are at right angles 22 times.
∴ In 24 hours, they are at right angles 44 times.

Question: {(2.39)² - (1.61)²}/(2.39 - 1.61) = ?

Solution: 
Let, 2.39 = a 
And 1.61 = b

Now, 
(a² - b²)/(a - b)
= (a + b)(a - b)/(a - b)
= a + b
= 2.39 + 1.61
= 4

10% of X = 20% of Y
or, (10/100) × X = (20/100) × Y
or, X/10 = Y/5
or, 5X = 10Y
or, X = 2Y
or, X/Y = 2/1
So, X : Y = 2 : 1

CP of 1 orange
= Tk.34/8
= Tk.4.25
SP of 1 orange
=Tk. 57/12
=Tk.4.75
Profit on each apple = (4.75 - 4.25) = Tk. 0.50
∴ Number of apples required
= 45/ 0.50
=90